Chapter 4 : Bonds in solid

Example 4.1 Page No : 137

In [1]:
			
import math 

# Variables
r_o = 2.8			#interatomic distance in Å
R_o = 2.8*10**(-10);#interatomic distance in m
u_o = 8.;			#released energy in eV
e = 1.6*10**(-19);	#charge of electron in C
U_o = 8.*e			#released energy in Joule

# Calculation
A = (5./4)*U_o*(R_o**2);			#proportionality constant for attraction in J-m2
B = A*(R_o**8)/5;			#proportionality constant for repulsion in J-m2
r_c = (110*B/(6*A))**(1./8);			#interatomic distance at which the dissociation occurs in m
F = -(2/r_c**3)*(A-5*B/(r_c**8));			#the force required to dissociate the molecule in N

# Results
print 'proportionality constant for attraction = %.2e J-m2'%A
print 'proportionality constant for repulsion = %.2e J-m2'%B
print 'interatomic distance at which the dissociation occurs = %.2e m'%r_c
print 'the force required to dissociate the molecule = %.2e N'%F

proportionality constant for attraction = 1.25e-37 J-m2
proportionality constant for repulsion = 9.48e-115 J-m2
interatomic distance at which the dissociation occurs = 3.29e-10 m
the force required to dissociate the molecule = -5.11e-09 N

Example 4.2 Page No : 138

In [2]:
			
import math 

# Variables
r_o = 3.14;			#nearest neighbour equilibrium distance in Å
R_o = 3.14*10**(-10);			#nearest neighbour equilibrium distance in m
K = 5.747*10**(-11);			#compressibility of KCl in m2/N
M = 1.748;			#Madelung constant
pi = 22./7;

# Calculation
E_o = 8.854*10**(-12);
q = 1.6*10**(-19);			#electron charge
n = 1+18*(R_o**4)*4*pi*E_o/(K*M*q**2);

# Results
print 'repulsive exponent n = %.1f'%n

repulsive exponent n = 8.6

Example 4.3 Page No : 139

In [4]:
			
import math 

# Variables
F_1 = 3.02*10**(-9);			#force of attraction b/w ions of Na+ and Cl-
Z_1 = +1;
Z_2 = -1;
e = 1.6*10**(-19);
E_o = 8.854*10**-12;
pi = 22./7;
r_Na = 0.95;			#ionic radius of Na+ ion

# Calculation
r = (-Z_1*Z_2*e**2/(4*pi*E_o*F_1))**(1./2);			#Radius of ion in meter
R = r/10**(-10);			#Radius of ion in Angstrom
r_Cl = (R-r_Na);			#Radius of Cl- ion in Angstrom

# Results
print 'Ionic Radius of Cl- ion in = %.2f Angstrom'%r_Cl

Ionic Radius of Cl- ion in = 1.81 Angstrom

Example 4.4 Page No : 139

In [5]:
			
import math 

# Variables
Z_1 = +2;
Z_2 = -2;
r_Mg = 0.65;			#radius of Mg++ ion
r_S = 1.84;			#radius of S-- ion
r = r_Mg+r_S;			#net radius(in Angstrom)

# Calculation
R = r*10**(-10);			#net radius(in meter)
e = 1.6*10**(-19);
E_o = 8.854*10**-12;
pi = 22./7;
F = -Z_1*Z_2*e**2/(4*pi*E_o*R**2);			#force of attraction between ions(in Newton)

# Results
print 'force of attraction between ions in = %.1e Newton'%F

force of attraction between ions in = 1.5e-08 Newton

Example 4.5 Page No : 150

In [7]:
			
import math 

# Variables
#Na atom requires +5.14 eV of energy. When this electron is transferred to a vacant position,it gives back –4.02 eV of energy
E_1 = +5.14;			#in eV
E_2 = -4.02;			#in eV

# Calculation
NET_energy = E_1+E_2;			#in eV

# Results
print 'Net spent energy in whole process in = %.2f eV'%NET_energy

Net spent energy in whole process in = 1.12 eV

Example 4.6 Page No : 150

In [8]:
			
import math 

# Variables
Enthalpy = 6.02;			#enthalpy of fusion of ice is 6.02 kJ/mol
E_h = 20.5;			#Hydrogen bond energy (in kJ/mol)
#There are two moles of hydrogen bonds per mole of H2O in ice.

# Calculation
H_b = Enthalpy/(2*E_h);			#the fraction of hydrogen bonds that are broken when ice melts

# Results
print 'fraction of hydrogen bonds that are broken when ice melts = %.2f'%H_b

fraction of hydrogen bonds that are broken when ice melts = 0.15